Let \(M = ([0, 1] \times [0, 1]) / \sim\) be the standard Möbius strip, where \((0, y) \sim (1, 1 - y)\). (1) Prove that \(M\) deformation-retracts onto its central circle \(C = \{[(x, 1/2)] \mid x \in [0, 1]\} \cong S^1\). (2) Compute the fundamental group \(\pi_1(M, x_0) \cong \mathbb{Z}\) and identify its generator \([\alpha]\). (3) Describe the boundary curve \(\partial M\) and prove that the inclusion \(\iota: \partial M \hookrightarrow M\) induces the map \(\iota_*: \pi_1(\partial M) \to \pi_1(M)\) given by multiplication by \(2\) (\([\partial M] = \alpha^2 \in \pi_1(M)\)).
Solution
Let \(q:[0,1]\times[0,1]\to M\) be the quotient map.
1. Define\[ H_s(q(x,y))=q\!\left(x,(1-s)y+s/2\right). \]This is well-defined because\[ (0,y)\sim(1,1-y) \]implies\[ \left(0,(1-s)y+s/2\right) \sim \left(1,1-((1-s)y+s/2)\right) = \left(1,(1-s)(1-y)+s/2\right). \]Thus \(H\) descends to \(M\), fixes the core\[ C=\{q(x,1/2):0\le x\le1\}, \]and at \(s=1\) maps all of \(M\) onto \(C\). Hence \(C\) is a strong deformation retract of \(M\).
2. Since \(C\cong S^1\),\[ \pi_1(M)\cong\pi_1(C)\cong\mathbb Z. \]With basepoint \(q(0,1/2)\), let\[ \alpha(t)=q(t,1/2),\qquad 0\le t\le1. \]Then \([\alpha]\) is a generator.
3. The boundary of \(M\) is one circle. A positively chosen parametrization can be written as the concatenation of\[ \gamma_0(t)=q(t,0),\qquad \gamma_1(t)=q(t,1), \]where the endpoint of \(\gamma_0\) satisfies\[ q(1,0)=q(0,1), \]and the endpoint of \(\gamma_1\) satisfies\[ q(1,1)=q(0,0). \]Thus \(\gamma_0*\gamma_1\) traverses the whole boundary once.
4. Under the deformation retraction to \(C\), both \(\gamma_0\) and \(\gamma_1\) become the core loop \(\alpha\) with the same orientation. Therefore\[ [\partial M]=[\alpha]^2\in\pi_1(M). \]Equivalently, after identifying both fundamental groups with \(\mathbb Z\) using these orientations,\[ \iota_*:\pi_1(\partial M)\longrightarrow\pi_1(M) \]is multiplication by \(2\).